YES
by ttt2 (version ttt2 1.15)
The rewrite relation of the following TRS is considered.
| a(x0) | → | x0 | 
| a(b(x0)) | → | c(b(c(b(x0)))) | 
| b(x0) | → | a(a(x0)) | 
| c(c(x0)) | → | x0 | 
| a#(b(x0)) | → | c#(b(x0)) | 
| a#(b(x0)) | → | b#(c(b(x0))) | 
| a#(b(x0)) | → | c#(b(c(b(x0)))) | 
| b#(x0) | → | a#(x0) | 
| b#(x0) | → | a#(a(x0)) | 
The dependency pairs are split into 1 component.
| b#(x0) | → | a#(a(x0)) | 
| a#(b(x0)) | → | b#(c(b(x0))) | 
| b#(x0) | → | a#(x0) | 
| [b(x1)] | = | 
 
  | 
||||||||
| [a#(x1)] | = | 
 
  | 
||||||||
| [b#(x1)] | = | 
 
  | 
||||||||
| [a(x1)] | = | 
 
  | 
||||||||
| [c(x1)] | = | 
 
  | 
| a(x0) | → | x0 | 
| a(b(x0)) | → | c(b(c(b(x0)))) | 
| b(x0) | → | a(a(x0)) | 
| c(c(x0)) | → | x0 | 
| a#(b(x0)) | → | b#(c(b(x0))) | 
The dependency pairs are split into 0 components.